How did Ramanujan get this result? We know Ramanujan got this result
$$\sqrt{1+2\sqrt{1+3\sqrt{1+\cdots }}}=3$$
and he used the formula
$$x+n+a=\sqrt{ax+{{(n+a)}^{2}}+x\sqrt{a(x+n)+{{(n+a)}^{2}}+(x+n)\sqrt{\cdots }}}$$
where $x=2,n=1,a=0$ ,we get the first result, but I don't know how to prove it, can you help me? 
 A: $$(x+n+a)^2 = x^2 + n^2 + a^2 + 2an + 2ax + 2nx$$ $$ = ax + (n+a)^2 + x(x + a + 2n)$$
so $x + n + a = \sqrt{ax + (n+a)^2 + x*((x+n) + n + a)}$
which you can substitute for $(x+n) + n + a$ again and again to get the sequence of iterated roots.
The convergence is because the sequence is monotone increasing but bounded above by $x+n+a$ for $n > 0$, $a,x \ge 0$ (after enough ($k$) iterations, $x + k*n + a$ will be greater than 1.)
A: Just use the following formula repeatedly
$ n^2=1+(n-1) (n+1) $
then we have
$ 2=\sqrt{1+1\times 3} $
$ 2=\sqrt{1 +1 \sqrt{1+2\times 4}}$
$ 2=\sqrt{1 +1 \sqrt{1+2 \sqrt{1+3\times 5}}} $
$ 2=\sqrt{1 +1 \sqrt{1+ 2 \sqrt{1+ 3 \sqrt{1+4\times 6}}}} $
$ 2=\sqrt{1+1 \sqrt{1+2 \sqrt{1+ 3 \sqrt{1+4 \sqrt{1+5\times 7}}}}}$
$ 2=\sqrt{1+1 \sqrt{1+2 \sqrt{1+3 \sqrt{1+4 \sqrt{1+5 \sqrt{1+6\times 8}}}}}} $
$ \cdots $
By using
$ n+(n-1) n(n+1) =n^3 $
I found the below new results
$ \sqrt[3]{2+1\times2 \sqrt[3]{3+2\times 3 \sqrt[3]{4+3\times 4 \sqrt[3]{5+4\times 5 \sqrt[3]{6+5\times 6 \sqrt[3]{7+...\ }}}}}}=2 $
$  \sqrt[3]{3+2\times 3 \sqrt[3]{4+3\times 4 \sqrt[3]{5+4\times 5 \sqrt[3]{6+5\times 6 \sqrt[3]{7+6\times 7 \sqrt[3]{8+...\ }}}}}}=3 $
$  \sqrt[3]{4+3\times 4 \sqrt[3]{5+4\times 5 \sqrt[3]{6+5\times 6 \sqrt[3]{7+6\times 7 \sqrt[3]{8+7\times 8 \sqrt[3]{9+...\ }}}}}}=4 $
$ \cdots $
More result can be found below:
http://eslpower.org/Notebook.htm
